See the opposite wall on a mirror I'm in tenth standard. This is a higher-order-thinking-skills Q I found in a book. One is supposed to use laws of reflection ($\angle i = \angle r$). You can also use mathematical concepts like similarity and trigonometry.

You are in the centre of a room (I assume cuboidal). There is a plane
  mirror hanging on one wall. If your eyes are $h$ distance from the
  ground, what is the minimum height of the plane mirror required to see
  the entire wall behind you.

Initially, I didn't get the answer. The answer in the book says $h/2$.
Then I came up with a solution that said:
$$Mirror\ height=\frac{Wall\ height}3$$
and I thought that the mirror height was independent of $h$.
What is the answer?
I just had a thought. Maybe we are allowed to turn the mirror to an angle. But that is  not clearly specified and I don't know how it would benefit us.
 A: To see complete wall (of height $l$) behind himself, a person requires a plane mirror of at least $\frac{1}{3}$ of the height of wall. It should be noted, as mentioned in the problem, that person is standing in the middle of the room.

The ray diagram is more or less like this.
The next diagram is however not that good . But for explanation consider it to work good.

From picture, we proceed to prove as follows:
Say $$MM' = r \,\ unit \,\ and \,\  EA = EG = HF = BD = MM' = r .....(i) \,\ (\,\ by \,\ construction)$$
In $\Delta$ RMB, C is mid-pt. of RB and CA is parallel to RM.
So A is mid-pt. of BM.
Similarly, I is mid-pt. of QH implies that G is mid-pt. of HM'.
Again in $\Delta$ MFB, A is mid-pt. of BM and AE is parallel to BF.
So, E is mid-pt. of MF and from the mid-pt. theorem on $\Delta$ M'DH, E is mid-pt. of M`D.
Therefore, $$2EA = FB \,\ and \,\ 2GE = HD \,\ implies \,\ HD = FB = 2r ....(ii)$$
But from (i) and (ii), we can say that $$HD = HF + FD = r + FD = 2r$$ implies $$FD = r$$
Therefore $$HF = FD = BD = r \,\ i.e. \,\  HB = HF + FD + BD = 3r$$
Therefore $$l = 3r$$ implies $$r = \frac{l}{3}$$
EDIT: Now according to the question, the person's eyes are at a height $h$ from ground.
So, $$EA = r \,\ and \,\ AC = h-r$$
And from $\Delta$ BRM, $$MR = 2AC = 2(h-r)$$
Similarly M'Q = 2(h-r)
Now as assumed HB = QR = l,
$$MR + MM' + M'Q = QR = l \,\ implies \,\ 2(h-r) + r + 2(h-r) = l$$
or, $$4h - 3r = l$$
And as earlier proved $r = \frac{l}{3}$
we have $$ 4h - 3r = 3r \,\ implies \,\ 6r = 4h \,\ or, \,\ r = \frac{2h}{3} $$
So your answer is $\frac{2}{3}$rd of the height at which the eye is stationed.
Hope it helps you.
A: The answer is $ \dfrac{L}{2} $, where $L$ is the length and height of the wall. When you are standing in such a way that your eyes are at height $h$, the height of the wall above you is $L-h$. Now to see that part, you need a  mirror of height $\dfrac{L-h}{2} $ because of $ \textit{i=r}$, you will be able to see rest $\dfrac{L-h}{2}$ . By the same logic, the mirror needs to have $\dfrac{h}{2}$ below the level of eyes. Thus, height of mirror must be $ \dfrac{L-h}{2} + \dfrac{h}{2} = \dfrac{L}{2}$. Same logic applies to the other dimension of the mirror. So, if it is a $ AXB$ wall , then you need $\dfrac{A}{2} X \dfrac{B}{2}$ mirror to watch the wall at your back.
